Question

$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+........(n$ terms$)=$?

• A. $1+\frac{x^n}{(x+1)(x+2)(x+3)..(x-n)}$
• B. $1-\frac{x^n}{(x+1)(x+2)(x+3)..(x+n)}$
• C. $1+\frac{x^n}{(x-1)(x-2)(x-3)..(x-n)}$
• D. $1-\frac{x^n}{(x+1)(x+2)(x+3)..(x-n)}$

Answer 1

The correct option is B $1-\frac{x^n}{(x+1)(x+2)(x+3)..(x+n)}$

Given

$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+........$ (n terms)

$=1-\frac{x}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+......$ (n terms)

$=1-\frac{x}{x+1}(1-\frac{2}{x+2})+\frac{3x^2}{(x+1)(x+2)(x+3)}+......$ (n terms)

$=1-\frac{x^2}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+......$ (n terms)

$=1-\frac{x^2}{(x+1)(x+2)}(1-\frac{3}{x+3})+......$ (n terms)

$=1-\frac{x^3}{(x+1)(x+2)(x+3)}+......$ (n terms)

Continuing in the same way, we get

$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+........(n$ terms$)$

$=1-\frac{x^n}{(x+1)(x+2)(x+3)..(x+n)}$